RVHS H2 CHEM P3 Ans Prelim

River Valley High School 9647/03/PRELIM II/16 [Turn over 2016 Preliminary Examination II RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION II CANDIDATE NAME CLASS 6 CENTRE NUMBER S INDEX NUMBER H2 CHEMISTRY 9647/03 Paper 3 Free Response 21 September 2016 2 hours Candidates answer on separate paper. Additional Materials: Answer Paper Cover Page Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class, centre number and index number on all the work you hand in. Write in dark blue or black pen on both sides of paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer any four questions. Begin each question on a fresh sheet of paper. A Data Booklet is provided. Do not write anything on it. You are reminded of the need for good English and clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together, with the cover page on top.

2 River Valley High School 9647/03/PRELIM II/16 2016 Preliminary Examination II Answer any four questions. 1 Many compounds of chlorine are manufactured from brine, NaC l(aq). The electrolysis of brine produces Cl2(g) and NaOH(aq). In some industrial electrolytic cells, these two substances are allowed to r eact further. The products formed in this second reaction depends on the operating conditions used. (a) Write balanced equations for the reaction between Cl2(g) and (i) cold aqueous NaOH; [1] Cl2 + 2NaOH  NaCl + NaClO + H2O (ii) hot aqueous NaOH. [1] 3Cl2 + 6NaOH  5NaCl + NaClO3 + 3H2O (b) Chlorine dioxide, ClO2, is used in the treatment of water. It is formed from ClO3− ions in an acidic solution. ClO3− + 2H+ + e− ⇌ ClO2 + H2O (i) Draw the dot-and-cross diagrams of ClO2 and H2O2. You may assume that there is no dative bond in either compound. [2] (ii) Construct the overall equation for the reaction of C lO3– ions with H2O2 in acidic solution. [1] 2ClO3 + 2H+ + H2O2  2ClO2 + 2H2O + O2 (iii) What is the role of H2O2 in the reaction? [1] Reducing agent

3 River Valley High School 9647/03/PRELIM II/16 [Turn over 2016 Preliminary Examination II (c) Chlorine can also be used to disinfect water. When chlorine is added to water, it produces hypochlorous acid, HOCl. Hypochlorous acid is a weak acid that dissociates into hypochlorite ions , OC l−, according to the following equation. HOCl(aq) ⇌ H+(aq) + OCl−(aq) Free chlorine refers to the total chlorine content in HOCl and OCl−. The dissociation curve below illustrates the ratio of hypochlorous acid to hypochlorite at different pH values. [Source: www.hach.com] (i) Determine the pKa of hypochlorous acid [1] pH = pKa when [HA] = [A −]. From the graph, pH = 7.5 when [OCl−] = [HOCl]. Therefore, pKa is 7.4 – 7.6. (ii) Hence, calculate the pH of a 0.0025 mol dm –3 hypochlorous acid solution. [2] pKa = 7.5, Ka = 10−7.5 = 3.16 × 10−8 mol dm−3 [H+] = (3.16 × 10−8 × 0.0025)1/2 = 8.89 × 10−6 mol dm−3 pH = − lg (8.89 × 10−6) = 5.05 To determine whether the free chlorine in a sample of tap water meets the regulatory limit (4 mg Cl per litre), OCl – is quantitatively reduced to Cl − by I− ions, which is in turn oxidised to I2. The I2 is titrated with standard sodium thiosulfate. The following reaction takes place during the titration. I2 + 2S2O32− → 2I− + S4O62− (iii) Construct a balanced equation between OCl–(aq) and acidified KI(aq) [1] 2H+(aq) + 2I−(aq) + OCl– (aq) → Cl– (aq) + I2(aq) + H2O(l) OCl– HOCl

4 River Valley High School 9647/03/PRELIM II/16 2016 Preliminary Examination II (iv) When 2 dm 3 of tap water was tested, 6.0 cm 3 of 0.00455 mol dm−3 sodium thiosulfate was required to discharge the colour of iodine. Calculate the concentration of OCl− in the sample of water. [2] Amount of S2O32− = 2.73 × 10−5 mol Amount of iodine produced = 1.37 × 10−5 mol Amount of OCl− present = 1.37 × 10−5 mol Concentration = 6.83 × 10−4 mol dm−6 (v) Suggest why the calculated concentration of OC l− has the same value as the concentration of free chlorine. [2] In the equilibrium , HOC l(aq) ⇌ H+(aq) + OC l−(aq), as OC l− is reduced by I−, [OCl−] decreases and the eq uilibrium position shifts right to favour the formation of more OC l− until all the HOC l completely dissociates. (vi) Determine if the sample of tap water is safe for consumption. [2] Mass concentration of free chlorine = 6.83 × 10 −6 × 35.5 = 0.242 mg per litre. The tap water is safe for consumption. (d) Chlorine is also used in organic chemistry to produce the Lewis acid catalyst, FeCl3, for the reaction between methylbenzene and chlorine. (i) Describe the mechanism of the above reaction. [3] FeCl3 + Cl2  FeCl4 + Cl+ (ii) FeCl3 reacts in a similar way with acyl chlorides. Predict the structure of the product of the following reaction. [1]

5 River Valley High School 9647/03/PRELIM II/16 [Turn over 2016 Preliminary Examination II [Total: 20]

6 River Valley High School 9647/03/PRELIM II/16 2016 Preliminary Examination II 2 (a) Nitrous oxide or dinitrogen oxide, N 2O, is commonly known as "laughing gas" due to the euphoric effects of inhaling it. It is used in surgery and dentistry for its anesthetic and analgesic effects. To produce N 2O, ammonium nitrate is decomposed at 170 C. Water is a by-product of this reaction. (i) Write an equation for the production of dinitrogen oxide from ammonium nitrate. [1] NH4NO3 (s) → 2 H2O (g) + N2O (g) (ii) In the manufacturing of N 2O gas, 1 kg of N 2O gas is produced for every 2.1 kg of ammonium nitrate used. Determine the percentage yield of dinitrogen oxide. [2] Amount of NH4NO3 used = 3164214 10001.2   = 26.3 mol Theoretical amt of N2O produced = 26.3 mol Theoretical mass of N2O produced = 26.3 × (14 × 2+ 16) = 1157g % yield = 1157 1000 × 100% = 86.4% (b) At 1200 K, in the presence of gold wire, dinitrogen oxide decomposes as shown: 2N2O(g)  2N2(g) + O2(g). To follow the rate of reaction, the change in concentration of a sample of N2O is measure against time. The results are shown below: Time, t / s Concentration of N 2O / × 10 3 mol dm3 0 2.50 1000 2.01 2000 1.62 3000 1.31 4000 1.05 5000 0.85 6000 0.68 7000 0.55

7 River Valley High School 9647/03/PRELIM II/16 [Turn over 2016 Preliminary Examination II (i) What do you understand by the term half-life of N2O? [1] Half-life is time taken for the concentration of N2O to be halved. (ii) Plot the above data on a graph paper. Use the following scale:  2 cm to represent 1000 s on the x-axis; and  2 cm to represent to represent 0. 25 × 103 mol dm3 on the y-axis. [2] (iii) From your graph, deduce the order of the reaction with respect to N2O. [2] When concentration decreases from 2.5 × 103 to 1.25 × 103, time taken is 3300 s. When concentration decreases from 1.25 × 103 to 0.7 × 103, time taken is 3100 s. (to be shown on the graph) Since half-life is approximately constant, the order of reaction with respect to N2O is 1. Show on graph that half time constant (iv) Calculate the rate constant for the reaction and state its units. [2] Rate constant = 2/1 2ln t = 2.17 x 104 s1 Correct answer Correct units (v) The gold wire acts as a heterogeneous catalyst in this reaction. [3] 0 0.5 1 1.5 2 2.5 3 0 2000 4000 6000 8000 Concentration of N2O / × 103 mol dm3 Time/s

8 River Valley High School 9647/03/PRELIM II/16 2016 Preliminary Examination II Explain the terms in italics and outline the mode of action of the catalys